A photoelectric surface is illuminated succes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$, where $\ph

Vedclass · Accepted Answer

$\frac{hc}{3\lambda}$

Vedclass · Answer

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$, where $\phi$ is the work function.For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.For the second case with wavelength $\lambda/3$: $K_2 = \frac{hc}{(\lambda/3)} - \phi = \frac{3hc}{\lambda} - \phi$.Given that $K_2 = 4K_1$, we substitute the expressions:$\frac{3hc}{\lambda} - \phi = 4(\frac{hc}{\lambda} - \phi)$.$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$.Rearranging the terms to solve for $\phi$:$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$.$3\phi = \frac{hc}{\lambda}$.$\phi = \frac{hc}{3\lambda}$.

Similar Questions

The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

$A$ sodium surface is illuminated by light of wavelength $3000\ \mathring A$. The work function of sodium is $2.6\ eV$. The maximum $K.E.$ of the emitted electrons is ........ $eV$.

Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about:

Vedclass Test Series

Exam Paper Generator

Online Exam Module